Giúp mị với mọi ngừi ơi:(((((((

Giải thích các bước giải:

 ĐK: $x,y,z>0$

Ta có:

$\begin{array}{l}
x + y + z + xy + yz + xz = 6xyz\\
 \Leftrightarrow \dfrac{{x + y + z + xy + yz + xz}}{{xyz}} = 6\\
 \Leftrightarrow \dfrac{1}{{yz}} + \dfrac{1}{{xz}} + \dfrac{1}{{xy}} + \dfrac{1}{z} + \dfrac{1}{x} + \dfrac{1}{y} = 6
\end{array}$

Mà lại có:

$\begin{array}{l}
A = \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} + \dfrac{1}{{{z^2}}}\\
 \Rightarrow 3A + 3 = \left( {\dfrac{1}{{{x^2}}} + 1} \right) + \left( {\dfrac{1}{{{y^2}}} + 1} \right) + \left( {\dfrac{1}{{{z^2}}} + 1} \right) + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}}} \right) + \left( {\dfrac{1}{{{y^2}}} + \dfrac{1}{{{z^2}}}} \right) + \left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{z^2}}}} \right)\\
 \Rightarrow 3A + 3 \ge 2\sqrt {\dfrac{1}{{{x^2}}}}  + 2\sqrt {\dfrac{1}{{{y^2}}}}  + 2\sqrt {\dfrac{1}{{{z^2}}}}  + 2\sqrt {\dfrac{1}{{{x^2}}}.\dfrac{1}{{{y^2}}}}  + 2\sqrt {\dfrac{1}{{{y^2}}}.\dfrac{1}{{{z^2}}}}  + 2\sqrt {\dfrac{1}{{{x^2}}}.\dfrac{1}{{{z^2}}}} \\
 \Rightarrow 3A + 3 \ge 2.\dfrac{1}{x} + 2.\dfrac{1}{y} + 2.\dfrac{1}{z} + 2.\dfrac{1}{{xy}} + 2.\dfrac{1}{{yz}} + 2.\dfrac{1}{{xz}}\\
 \Rightarrow 3A + 3 \ge 2\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} + \dfrac{1}{{xy}} + \dfrac{1}{{yz}} + \dfrac{1}{{xz}}} \right)\\
 \Rightarrow 3A + 3 \ge 2.6\\
 \Rightarrow A \ge 3
\end{array}$

Dấu bằng xảy ra

$ \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} = 1\\
\dfrac{1}{y} = 1\\
\dfrac{1}{z} = 1\\
\dfrac{1}{x} = \dfrac{1}{y}\\
\dfrac{1}{y} = \dfrac{1}{z}\\
\dfrac{1}{x} = \dfrac{1}{z}
\end{array} \right. \Leftrightarrow x = y = z = 1$

Vậy ta có điều phải chứng minh.